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When does a BJT conduct electricity and from where to where does it flow?

I thought a BJT is off when the base is low and otherwise the electricity flows from emitter to collector (pnp) or from collector to emitter (npn), right?

I have this exercise with solution for example. How can I figure out if a BJT is on or off?

For T3 it seem's to be simple: It's on when A is high because then T3's base is high.

But why is T7 always on although it's connected to ground?

How can T9 be on although it's base is connected with T4's collector?

Exercise Solution: Solution[2]

Seem's like I didn't understand how BJTs work overall...

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  • \$\begingroup\$ "Here's an ECL gate. Learn how BJTs work by studying its operation." Wow. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 5 '15 at 8:43
  • \$\begingroup\$ Note here, the POSITIVE rail is GND, and the voltage supply is labelled -UB (usually -5.1V) "U" for Voltage is typically used in Germany. But of all the logic families in existence, I only know of one that uses -ve supply, that is ECL (Emitter Coupled Logic) as @IgnacioVazquez-Abrams says. Fastest thing around in the 1970s (compare the bandwidth of an emitter follower with that of a common emitter amplifier to see why) but a serious oddball today... NOT the plate to start learning... \$\endgroup\$ – Brian Drummond Sep 5 '15 at 10:11
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I agree with Brian and Ignacio that this (an ECL exlusive-or gate) is a bit much to analyze for a beginner.

T7 and T8 (and the associated diodes and resistors) establish bias voltages. You can assume UH1 and UH2 are constant voltages between '0' and '1'. This is one missing part to the puzzle.

The other important thing to know is that the transistor with the higher base voltage in a differential pair will be 'on' when its base voltage is higher than the base voltage of its counterpart (assuming it's seeing some emitter current through a resistor or current sink- a resistor in this case).

So right away we can see that only one pair of T3/T4 or T5/T6 will get emitter current, depending on whether B is higher or lower than UH2. The other pair will have no effect.

So if B is high, then T3/T4 are active, and if A is high then T3 is on. If T3 is on, then T10 is off and Q2 is low. T4 is off and Q1 is high.

You can figure out the other three possibilities.

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